Question

The equations of the diagonals of the square formed by the pairs of straight lines 3x^2 + 8xy - 3y^2 = 0 and
3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0 are
(1) x = 2y, 4x + 2y + 1 = 0
(2) 2x + y = 0, 2x = 4y + 1
(3) x = 2y, 2x = 4y + 1
(4) 2x + y = 0, 4x + 2y + 1 = 0​

Answer 1

Given : Square formed by the pairs of straight lines 3x^2 + 8xy - 3y^2 = 0 and  3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0

To Find : The equations of the diagonals of the square

Solution:

3x² + 8xy - 3y²= 0

=> 3x² + 9xy - xy - 3y²= 0

=>3x(x + 3y) - y(x + 3y) = 0

=> (3x - y)(x + 3y) = 0

3x - y = 0

x + 3y = 0

3x² + 8xy - 3y² + 2x - 4y - 1  = 0

(3x - y - 1)(x+ 3y + 1)  = 0

3x - y - 1 = 0

x+ 3y + 1 = 0

Solving Pairs Points of intersections :  

 (-0.1 , -0.3)  , (0.2 , -0.4) , (0.3 , -0.1)  , (0 , 0)

Diagonal   (-0.1 , -0.3) (0.3 , -0.1)

(0.2 , -0.4)  , (0 , 0)  => y = -2x => 2x + y = 0  

(-0.1 , -0.3) (0.3 , -0.1)  => y + 0.3=  (1/2) (x + 0.1)

=> 2y + 0.6 = x + 0.1

=> 2y + 0.5 = x

=> 4y + 1 = 2x

2x + y = 0   and 2x =  4y + 1   are the Equation of Diagonals

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